<<PLAESE EXPLAIN HOW TO GET THE ANSWER>> A ball is thrown into the air at initia

Question

<<PLAESE EXPLAIN HOW TO GET THE ANSWER>>


A ball is thrown into the air at initial speed vo = 38.9 m/s, at an angle of ? = 40.7 o to the ground. The ball lands at the same elevation as it was thrown. Find:
- the horizontal distance the ball traveled: x = . m
- the maximum height the ball reached: hmax = . m

Explanation / Answer

Maximum height: v^2 = u^2+2as v = 0, u = 38.9sin(40.7) = 25.36662791m/sec, a = -g = -9.81m/sec^2,We have to find s =>0 = 25.36662791^2 - 2*9.81*hmax =>hmax = 32.7964226m Horizontal Distance: v = u+at v = 0, u = 38.9sin(40.7) = 25.36662791m/sec, a = -g = -9.81m/sec^2 =>0 = u-9.81t =>t = 2.585792855sec =>2t = 5.17158571sec x = 38.9*cos(40.7)*2t = 152.5174356m

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